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Byju's Answer
Standard X
Physics
Faraday's Law
Question 5 ...
Question
Question 5
5
.
A
n
o
n
c
o
n
d
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g
i
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f
i
n
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o
d
i
s
p
l
a
c
e
d
a
l
o
n
g
t
h
e
z
-
a
x
i
s
:
t
h
e
u
p
p
e
r
h
a
l
f
o
f
t
h
e
r
o
d
(
l
y
i
n
g
a
l
o
n
g
z
≥
0
)
i
s
c
h
a
r
g
e
d
p
o
s
i
t
i
v
e
l
y
w
i
t
h
a
u
n
i
f
o
r
m
l
i
n
e
a
r
c
h
a
r
g
e
d
e
n
s
i
t
y
+
λ
w
h
i
l
e
t
h
e
l
o
w
e
r
h
a
l
f
(
z
<
0
)
i
s
c
h
a
r
g
e
d
n
e
g
a
t
i
v
e
l
y
w
i
t
h
a
u
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f
o
r
m
l
i
n
e
a
r
c
h
a
r
g
e
d
e
n
s
i
t
y
-
λ
.
T
h
e
o
r
i
g
i
n
i
s
l
o
c
a
t
e
d
a
t
t
h
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j
u
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f
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p
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e
g
a
t
i
v
e
h
a
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v
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s
o
f
t
h
e
r
o
d
.
A
u
n
i
f
o
r
m
l
y
c
h
a
r
g
e
d
a
n
n
u
l
a
r
d
i
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c
(
s
u
r
f
a
c
e
c
h
a
r
g
e
d
e
n
s
i
t
y
:
σ
0
)
of
inner
radius
R
and
outer
radius
2
R
is
placed
in
the
x
-
y
plane
with
its
centre
of
the
origin
.
The
force
on
the
rod
due
to
the
disc
is
:
-
(
A
)
2
σ
0
λR
ε
0
(
B
)
σ
0
λR
2
ε
0
(
C
)
σ
0
λR
ε
0
(
D
)
σ
0
λR
3
ε
0
Open in App
Solution
D
e
a
r
s
t
u
d
e
n
t
D
e
a
r
s
t
u
d
e
n
t
Q
=
d
V
d
t
Q
=
d
π
r
2
h
d
t
d
h
d
t
=
Q
π
r
2
=
r
a
t
e
a
t
w
h
i
c
h
w
a
t
e
r
l
e
v
e
l
r
i
s
e
s
Suggest Corrections
0
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